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Homework Help: Partial Implicit Differentiation

  1. Jun 11, 2005 #1
    Just wondering if I did this right:

    Here is the question: find [tex]\frac{\partial z}{\partial x} of \frac{x^2}{9} - \frac{y^2}{4} + \frac{z^2}{2} = 1[/tex]

    Now I put the [tex] \frac{\partial z}{\partial x} [/tex] on both sides then got.

    [tex] \frac{2x}{9} - 0 + z \frac{\partial z}{\partial x} = 0 [/tex]

    So


    [tex] \frac{\partial z}{\partial x} = -\frac{2x}{9z} [/tex]

    Now I know this is the right answer I am just curious if I did it right, it has been a while.

    Now my reasoning:

    [tex] \frac{\partial z}{\partial x} \frac{x^2}{9} [/tex] is just the derivative with respect to x, so it will be [tex]\frac{2x}{9}[/tex]

    [tex] \frac{\partial z}{\partial x} \frac{y^2}{4} [/tex] has no z or x, so it is constand and therefore 0.

    [tex] \frac{\partial z}{\partial x} \frac{z^2}{2} [/tex] has a z, so it is the derivative of itself, but times [tex] \frac{\partial z}{\partial x} [/tex]

    Is all of that correct, or did I do something wrong. The book I have only shows one example :mad:

    Thanks!
     
    Last edited: Jun 11, 2005
  2. jcsd
  3. Jun 11, 2005 #2

    dextercioby

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    Homework Helper

    It looks okay.You could have done it using the theorem of implicit functions,as well.

    Daniel.
     
  4. Jun 11, 2005 #3
    Thanks dex! I have no idea what the theorem of implicit functions is, but I probably could not use it anyway.
     
  5. Jun 11, 2005 #4
    One more question for now :smile:

    I need to find

    [tex] \frac{\partial z}{\partial x} sin(xz) [/tex]

    How exactly does this work?

    Will it be: cos(xz) [times] what? [tex] \frac{\partial_}{\partial x} xz [/tex]

    Which will then be cos(xz) [times] x ? This one seems to be wrong. Any ideas?
     
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